Can particles traveling near the speed of light create a time reversal effect?

[John_M]

According to Einstein nothing can move faster than c. But whereas c is constant in all frames of reference, the speed of massive objects is defined relative to other massive objects. So when we say nothing can go faster than c, we surely mean nothing can go faster than c relative to another object.

So - I've heard of particles that travel quite close to c relative to the Earth. If I fire a particle like this in one direction, it will be traveling at a speed greater than 0.5 c relative to the Earth. If I then fire another particle of the same speed in the opposite direction, it will be traveling greater than 0.5c relative to the Earth and greater than c relative to the other particle. Correct?

Is it then correct to state that, in one of these particles' frame of reference, the other particle will be going backwards in time?

 

[chroot]

From your Earthly perspective, one particle will be moving at 0.5c in one direction, and the other will be traveling at 0.5c in the other direction.

From the particles' perspective, the Earth is moving at 0.5c, but the other particle is moving at

$$\frac{0.5c + 0.5c}{1 + \frac{(0.5c)^2}{c^2}} = 0.8c$$

Velocities in special relativity do not add in the normal Galilean way.

- Warren

 

[Nenad]

From an observer on earth, particles can be moving apart at 2c, but as long as this earthly observer is stationary and not in the motion frame of the paticles. Relative to the particle, you need to use relativistic velocity addition, like chroot said.

 

[John_M]

From your Earthly perspective, one particle will be moving at 0.5c in one direction, and the other will be traveling at 0.5c in the other direction.

From the particles' perspective, the Earth is moving at 0.5c, but the other particle is moving at

$$\frac{0.5c + 0.5c}{1 + \frac{(0.5c)^2}{c^2}} = 0.8c$$

Velocities in special relativity do not add in the normal Galilean way.

- Warren

How do they add then? It's news to me

I read something about 'neutral pions' going at 185,000 miles per second - presumably relative to the Earth - in experiments to test the speed of light. How does your statement that 'From your Earthly perspective, one particle will be moving at 0.5c in one direction, and the other will be traveling at 0.5c in the other direction' square with this? If you accept that it's possible for a particle to travel at a speed greater than 0.5c relative to the Earth (and it does seem to be) it seems odd to suggest that this particle will slow down to 0.5c relative to the Earth simply because another particle has been fired in the opposite direction. This is my problem (although I'm sure that's not what you're actually suggesting...)

Thanks

John.

 

[John_M]

actually, I could have read over your post a bit more closely...you've posted an equation for working out relativistic velocities...

 

[Janus]

How do they add then? It's news to me

it seems odd to suggest that this particle will slow down to 0.5c relative to the Earth simply because another particle has been fired in the opposite direction. This is my problem (although I'm sure that's not what you're actually suggesting...)



John.

No, this is not what he is suggesting. Both particles maintain a velocity of .05c relative to the Earth as measured by the Earth. Each particle also measures its relative velocity to the Earth as 0.5c. Each particle, however will measure the others particles relative velocity to itself as 0.8c (not 1c). if the particles relative velocity to the Earth were 0.75c, they would measure their respective relative velocity as 0.96c

 

[John_M]

"Both particles maintain a velocity of .05c relative to the Earth as measured by the Earth. Each particle also measures its relative velocity to the Earth as 0.5c."

It's this statement I don't understand. Do you accept that it's possible for a particle to travel at greater than 0.5c relative to the Earth and, if so, what's your basis for stating that 'both particles maintain a velocity of 0.5c relative to the Earth'? As I understand it, some particles are capable of traveling at 0.99c relative to the Earth - so where does the figure of 0.5c come from?

Thanks for the replies

John.

 

[chroot]

You're the one who brought up 0.5c. Bottom line is that no observer will ever see another object going faster than light.

An observer might see one object leaving at almost the speed of light in one direction, and another object leaving at almost the speed of light in another direction, but that doesn't violate that bottom line.

- Warren

 

[John_M]

"An observer might see one object leaving at almost the speed of light in one direction, and another object leaving at almost the speed of light in another direction, but that doesn't violate that bottom line."

Why not? It seems very clearly to violate the bottom line to me, seeing as relative speed is the only type of speed that exists in SR and there's no reason to prioritise the observer's frame over that of the objects. If what you say is true, doesn't it follow (if the objects are moving in directly opposite directions) that the first object will be moving at almost 2c relative to the second?

Tell me where I've gone wrong here. Assume all speeds are constant.

1) I fire a particle at a speed of 0.99c relative to the Earth.
2) I fire a second particle at a speed of 0.99c relative to the Earth in the opposite direction to the first particle.
3) Therefore Particle 1 is traveling at a speed of 1.98c relative to Particle 2.

Now, you might state that it's impossible to go faster than c, full stop, mass increases will happen etc etc. But saying 'it's impossible to go faster than c' - doesn't tell the whole story because it doesn't specify a frame.

If you accept that I can, at will, fire a particle at 0.99c - and in my limited understanding scientists have done just that - e.g. 'neutral pions' used to test the constancy of the speed of light - what exactly is stopping me from firing a second such particle in the opposite direction, and isn't it the case that the second particle will have a speed of 1.98c relative to the first?

 

[chroot]

Again, you're mixing reference frames. You can't assert that an observer on a moving particle will measure another moving particle the same way as a stationary observer between them.

If you fire two particles at 0.99 c from Earth in opposite directions, here are the relative speeds seen by all observers:

Particle 1:
Sees Earth moving at 0.99c
Sees Particle 2 moving at 0.999949498c

Earth:
Sees Particle 1 moving at 0.99c
Sees Particle 2 moving at 0.99c

Particle 2:
Sees Earth moving at 0.99c
Sees Particle 1 moving at 0.999949498c.

In no case is any speed > c.

- Warren

 

[John_M]

Where do you get the number 0.999949498c from?

 

[chroot]

From the way velocities add in special relativity, due to Lorentz transformations. I've already given you the form used in computing added velocities, but here it is again.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

- Warren

 

[jcsd]

John -

You're applying a Gallielan transformation, but what special relativity says is: "the Gallilean transformation is only an approximation that applies when the velcoites involved are much smaller than c". What you should be applying is a Lorentz transformation.

 

[John_M]

OK...thanks guys.

I think the statement from Chroot at the start clears it up...

'Velocities in special relativity do not add in the normal Galilean way.'

 

[Nenad]

Again, you're mixing reference frames. You can't assert that an observer on a moving particle will measure another moving particle the same way as a stationary observer between them.

If you fire two particles at 0.99 c from Earth in opposite directions, here are the relative speeds seen by all observers:

Particle 1:
Sees Earth moving at 0.99c
Sees Particle 2 moving at 0.999949498c

Earth:
Sees Particle 1 moving at 0.99c
Sees Particle 2 moving at 0.99c

Particle 2:
Sees Earth moving at 0.99c
Sees Particle 1 moving at 0.999949498c.

In no case is any speed > c.

- Warren

yes, the observer on Earth can see the two particles moving away from each other at 0.99c + 0.99c.

 

[LURCH]

John,
If you're looking for a more intuitive understanding, this is where time dilation comes in. Partical 1 sees Partical 2's movement through time as greatly slowed down. So Partical 2 may measure his speed as .99c relative to the Earth, but Partical 1 would say, "well sure he measures his own speed as 186,000 miles per second, but look at his watch; his seconds are almost twice as long as they should be!"

 

[robphy]

OK...thanks guys.

I think the statement from Chroot at the start clears it up...

'Velocities in special relativity do not add in the normal Galilean way.'

It is the "rapidities" (analogous to the Euclidean angles) that add.
Velocity is the hyperbolic-tangent of the rapidity.
The "addition [better: composition] of velocities" formula is a trigonometric identity involving the hyperbolic-tangent of the sum of two rapidities.

 

[jcsd]

And once you look the Lorentz transformation in terms of the rapidity, the simlairty between boost (a change in velocity) and rotation become obvious.

 

[robphy]

On top of that, calculations are much easier to execute and interpret because one can appeal to one's intuition on [hyperbolic-]trigonometric and exponential functions.

For the uninitiated,
$\gamma$, $\gamma v$, $v$ are $\cosh(\theta)$, $\sinh(\theta)$, and $\tanh(\theta)$.

Exercise: find the physical interpretation of $\exp(\theta)$. Hint: express $\exp(\theta)$ in terms of $v$.

 

[jcsd]

$$e^{\theta} = \cosh\theta + \sinh\theta = \gamma + \gamma v=$$
$$\gamma(1 + v) = \frac{1 + v}{\sqrt{1 - v^2}} =$$
$$\sqrt{\frac{(1 + v)^2}{1 - v^2}} = \sqrt{\frac{1 + v}{1 - v}} = \frac{x + t}{x' + t'}$$

latex is on the blink, the last terms should be (x + t)/(x' + t')

 

[selfAdjoint]

Thank heaven we've finally made it through to rapidities and hyperbolic trig functions! It's high time the smart inquisitive people we see here were introduced to this nifty methodology.

 

[robphy]

$$e^{\theta} = \cosh\theta + \sinh\theta = \gamma + \gamma v=$$
$$\gamma(1 + v) = \frac{1 + v}{\sqrt{1 - v^2}} =$$
$$\sqrt{\frac{(1 + v)^2}{1 - v^2}} = \sqrt{\frac{1 + v}{1 - v}} = \frac{x + t}{x' + t'}$$

latex is on the blink, the last terms should be (x + t)/(x' + t')

The last expression isn't necessary.
Physical interpretation of $$\sqrt{\frac{1 + v}{1 - v}}$$?


"An introduction to rapidities" might be a good thing.
I once asked Edwin Taylor why the use of rapidities was removed from the new version of Spacetime Physics. He said that many people didn't use it.

 

[jcsd]

The last expression isn't necessary.
Physical interpretation of $$\sqrt{\frac{1 + v}{1 - v}}$$?


"An introduction to rapidities" might be a good thing.
I once asked Edwin Taylor why the use of rapidities was removed from the new version of Spacetime Physics. He said that many people didn't use it.

I see now, it's the relativstic doppler shift, i.e. the ratio between the frequency of a light wave in two frames.

I'm not overly famliar with rapidites, but they do offer a very useful tool, as IMHO special relativity becomes more intutive when appraoched geometrically.

 

[DW]

I see now, it's the relativstic doppler shift, i.e. the ratio between the frequency of a light wave in two frames.

I'm not overly famliar with rapidites, but they do offer a very useful tool, as IMHO special relativity becomes more intutive when appraoched geometrically.

This sort of thing leads to an ineteresting correspondence between the equations of relativistic rocketry and Doppler shift. For example the relativistic rocket equation can be expressed as
$$v = ctanh(\frac{v_{ex}}{c}ln(\frac{m_{i}}{m}))$$
inverted
$$\frac{m_{i}}{m} = (\frac{c + v}{c - v})^{\frac{c}{2v_{ex}}}$$
And for the object traveling toward the observer Doppler shift can be expressed as
$$\frac{f}{f_{0}'} = (\frac{c + v}{c - v})^{\frac{1}{2}}$$
inverted
$$v = ctanh(ln(\frac{f}{f_{0}'}))$$
Its easy to do the inversions by starting with the tanh expressions, and then replace the tanh with its expression in terms of exp and then using exp(lnx) = x.

 

[robphy]

jcsd,
Yes, that is the Doppler factor.
Note that it follows immediately that $$\exp(-\theta)=\frac{1}{\exp(\theta)}=\sqrt{\frac{1-v}{1+v}}$$.
I should also point out that $\exp(\theta)=k$, as in Bondi's k-calculus... another simple but rarely taught approach to SR.


DW,
That's interesting. It might be enlightening to rederive that rocket result starting from the hyperbolic-trig expressions.

 

[HallsofIvy]

Why do you quote what Chroot did, then say "yes" and finish by contradicting what you had just quoted?

An observer on Earth can see an object moving away from himself at 0.99c and see an object moving awayfrom himself, in the opposite direction at 0.99c but athat doesn't mean that he can "see them moving away from each other at 0.99c+ 0.99c". He can't, in any reasonable sense, see them moving away from each other at a specific speed since that depends on the speed of one relative to the other and he can only calculate, not measure, that speed.

 

[DW]

jcsd,
Yes, that is the Doppler factor.
Note that it follows immediately that $$\exp(-\theta)=\frac{1}{\exp(\theta)}=\sqrt{\frac{1-v}{1+v}}$$.
I should also point out that $\exp(\theta)=k$, as in Bondi's k-calculus... another simple but rarely taught approach to SR.


DW,
That's interesting. It might be enlightening to rederive that rocket result starting from the hyperbolic-trig expressions.

Ok.
Conservation of energy:
$$mc^{2}cosh(\theta ) = (m + dm)(cosh(\theta ) + d(cosh(\theta )))c^{2} + m_{fex}\gamma _{fex}c^{2}$$
Conservation of momentum:
$$mcsinh(\theta ) = (m + dm)(sinh(\theta ) + d(sinh(\theta )))c + m_{fex}\gamma _{fex}ctanh(\theta _{fex})$$
Simplified
$$0 = sinh(\theta )dm +mcosh(\theta )d\theta + m_{fex}\gamma _{fex}tanh{\theta _{fex}}$$
$$0 = cosh(\theta )dm + msinh(\theta )d\theta m_{fex}\gamma _{fex}$$
Eliminate $$m_{fex}\gamma _{fex}$$
$$0 = sinh(\theta )dm + mcosh(\theta )d\theta - (cosh(\theta )dm + msinh(\theta )d\theta )tanh(\theta _{fex})$$
Insert rapidity addition $$\theta _{fex} = \theta - \theta _{ex}$$:
$$0 = sinh(\theta )dm + mcosh(\theta )d\theta - (cosh(\theta )dm + msinh(\theta )d\theta )tanh(\theta - \theta _{ex})$$
Using hyperbolic trig angle difference identity
$$0 = sinh(\theta )dm + mcosh(\theta )d\theta - (cosh(\theta )dm + msinh(\theta )d\theta )\frac{tanh(\theta ) - tanh(\theta _{ex})}{1 - tanh(\theta )tanh(\theta _{ex})}$$
This simplifies all the way down to
$$d\theta = - tanh(\theta _{ex})\frac{dm}{m}$$
Integration yields
$$\Delta \theta = tanh(\theta _{ex})ln(\frac{m_{i}}{m})$$
Expressed in terms of $$v_{ex}$$
$$\Delta \theta = \frac{v_{ex}}{c}ln(\frac{m_{i}}{m})$$
One then obtains the final velocity from
$$v = ctanh(\theta )$$
If one considers an initial velocity of zero then
$$v = ctanh(\frac{v_{ex}}{c}ln(\frac{m_{i}}{m}))$$

 

[Doc Al]

"separation rate" vs relative speed

yes, the observer on Earth can see the two particles moving away from each other at 0.99c + 0.99c.

Yes, if one particle moves at 0.99c to the left (with respect to the earth) and the other moves at 0.99c to the right, then an Earth observer will "see" them separate at 1.98c. I think it's best to call this a separation rate, rather than a "relative velocity" since neither particle is observed to move at that speed from any frame. The speeds of the particles with respect to each other are always less than c, as explained by Warren.

 

[Nenad]

I know, but I just wanted the readers to know that there can be things that are greater than c. But thanks for the correlation.